airyana1114
  • airyana1114
Can someone help me solve this equation? Step by Step.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
is it a hard equation ?
airyana1114
  • airyana1114
\[\sqrt{x+6}-4=x\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

misty1212
  • misty1212
oh no that can't be that hard
airyana1114
  • airyana1114
Can you help me? I've never had this type of equation equaled to x and it sort of makes me nervous. Can you walk me through it? @misty1212
misty1212
  • misty1212
yeah no problem first step is to get the radical by itself on one side of the equation by adding 4 to both sides, i.e. write \[\sqrt{x+6}=x+4\]
misty1212
  • misty1212
now you want to do it the math teacher way , or you want to think what the answer is , or both?
airyana1114
  • airyana1114
I'm not sure I understand what you're asking.
misty1212
  • misty1212
we can find the answer in our head, but that is probably not what your math teacher wants you to do ,so lets go ahead and solve it the way they do in math class
airyana1114
  • airyana1114
Okay
misty1212
  • misty1212
i can show you how to get the answer in your head second
airyana1114
  • airyana1114
Alright thanks :)
misty1212
  • misty1212
you want to get rid of the radical, so square both sides (carefully)
airyana1114
  • airyana1114
How would I square (x+4). Like this? x^2+8
misty1212
  • misty1212
\[\sqrt{x+6}=x+4\] square and get \[x+6=(x+4)^2\] or \[x+6=(x+4)(x+4)\] or \[x+6=x^2+8x+16\]
airyana1114
  • airyana1114
Where did the 8x come from?
misty1212
  • misty1212
no dear that is not how you square \(x+4\) you have to write it as \[x(x+4)(x+4)=x^2+4x+4x+16=x^2+8x+16\]
misty1212
  • misty1212
it is that old "first outer inner last" stuff
airyana1114
  • airyana1114
Ohhhh, FOIL okay..that makes sense.
misty1212
  • misty1212
so now we have \[x+6=x^2+8x+16\] which is a quadratic equation we have to solve
misty1212
  • misty1212
set it equal to zero, get \[x^2+7x+10=0\] then factor
misty1212
  • misty1212
\[x^2+7x+10=0\\ (x+2)(x+5)=0\] so \[x=-2,x=-5\] but we are not done yet
misty1212
  • misty1212
you have to check the answer in the original equation not the quadratic equation if you do, you will see that \(x=-2\) works but \(x=-5\) does not
airyana1114
  • airyana1114
Making it an extraneous solution correct?
misty1212
  • misty1212
right , which is a silly way of saying it is not a solution but no matter
airyana1114
  • airyana1114
:) thank you @misty1212
misty1212
  • misty1212
\[\color\magenta\heartsuit\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.